Two wires of same metal have the same length but their cross- sections are in the ratio 3 :1 . They are joined in series. The resistance of the thicker wire is `10 Omega` . The total resistance of the combination will be

Options

A. `5//2 Omega`
B. `40//3 Omega`
C. `40 Omega`
D. `100 Omega`

Answer

C

Explanation

Resistance of a wire `= rhol//A`.
For the same length and same material,
`(R_2)/(R_1)= (A_1)/(A_2) =(3)/(1)` or`, R_(2) = 3 R_(1)`
The resistance of thick wire, `R_1 = 10 Omega`
The resistance of thin wire `= 3 R_(1) = 3xx 10= 30 Omega`.
Total resistance `=10 + 30 = 40 Omega`.